## Precalculus (6th Edition) Blitzer

$(x+4)(x-4)(y-2)$
$x^{2}y-16y+32-2x^{2}=$... factor in pairs $=(x^{2}y-16y)+(32-2x^{2})=$ $=y(x^{2}-16)+2(16-x^{2})$... factor out -1 in the 2nd term $=y(x^{2}-16)-2(x^{2}-16)$ $=(x^{2}-16)(y-2)$ ...The first parentheses hold a difference of squares $(x)^{2}-(4)^{2}$ = $(x+4)(x-4)(y-2)$